This post contains problems from the first relay round of the Duke Math Meet 2009. Try to solve these problems.
1A. Find the lowest positive angle $ \theta $ that satisfies the equation $ \sqrt {1+\cos \theta} = \sin \theta + \cos\theta $ expressed in degrees.
Discussion:
$ \sqrt {1 +\cos\theta} = \cos\theta + \sin \theta \Rightarrow \sqrt{2\cos^2 \frac{\theta}{2} } = \sqrt2{\frac{1}{\sqrt2} \cos\theta + \frac{1}{\sqrt2} \sin\theta } $
Now this gives
$ \sqrt2 \cos\frac{\theta}{2} = \sqrt2\cos(\theta - \frac{\pi}{4}) \Rightarrow \frac{\theta}{2} = \theta - \frac{\pi}{4} $ or $ \frac{\theta}{2} = -\theta + \frac{\pi}{4}$
Thus the possible values of $ \theta $ are $ 90^o $ or $ 30^o $.
Since we require the smallest positive angle hence the answer is $ 30^o $.
1B. Let n be two times the tens digit of TNYWR. Find the coefficient of the $ x^{n-1}y^{n+1} $ term in the expansion of $ (2x + \frac{y}{2} + 3)^{2n} $
Discussion:
TNYWR is 3. Hence n = 6 Thus we are required to find coefficient of $ x^5 y^7 $ term in the expansion of $ (2x + \frac{y}{2} + 3 )^{12} $
This can be easily found from trinomial expansion. The required term is $ {{12}\choose {5}}(2x)^5 {{7}\choose{7}} (\frac{y}{2})^7 = 792 \times 32 \times \frac{1}{128} = 198 $
1C. Let k be TNYWR, and let n = k/2. Find the smallest integer m greater than n such that 15
divides m and 12 divides the number of positive integer factors of m.
Discussion:
k = 198, hence n = 99.
So we have to look at multiples of 15 greater than 99. We want 12 to divide the number of positive divisors of m.
Suppose $ m = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k} $. The number positive divisors of k is $ (\alpha_1 +1 )... (\alpha_k + 1) $
The first multiple of 15 greater than 99 is $ 105 = 15 \times 7 $ . By inspection we see that m = 150.
